Question

2y³-3y²+3y-1 (Solv it by using Vanishing Method)​

Answer 1

Step-by-step explanation:

Given that,

An expression 2y^3+y^2-2y-12y

3

+y

2

−2y−1

To find,

Factors of the above expression.

Solution,

We have,

2y^3+y^2-2y-12y

3

+y

2

−2y−1

Taking y² common from first two terms and -1 common from last two terms

\begin{gathered}=y^2(2y+1)-1(2y+1)\\\\=(y^2-1)(2y+1)\end{gathered}

=y

2

(2y+1)−1(2y+1)

=(y

2

−1)(2y+1)

Now using identity, (a^2-b^2)=(a-b)(a+b)(a

2

−b

2

)=(a−b)(a+b)

So,

=(y-1)(2y+1)(y+1)=(y−1)(2y+1)(y+1)

So, the factors of 2y^3+y^2-2y-12y

3

+y

2

−2y−1 is (y-1)(2y+1)(y+1)(y−1)(2y+1)(y+1) .

Answer 2

Step-by-step explanation:

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